Mesh Analysis for Circuit: Finding ix with Supermesh

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Discussion Overview

The discussion revolves around the application of mesh analysis in circuit analysis, specifically focusing on determining the current ix in a given circuit. Participants explore whether supermesh analysis is necessary in this context.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests that supermesh analysis is appropriate for finding ix.
  • Another participant counters that three loop equations with three unknowns should be written instead, noting that mesh analysis typically focuses on voltages and that supermesh is used when dealing with current sources.
  • A participant questions the relationship between the currents i3 and ix, asking if they are separate or if ix equals i3.
  • A later reply clarifies that while i3 and ix are different currents, ix can be expressed in terms of the mesh currents, specifically as I_X = I_2 - I_3.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the necessity of supermesh analysis for this problem, with differing views on the relationship between the currents involved.

Contextual Notes

There is an implicit assumption regarding the definitions of the currents and the circuit configuration, which may affect the application of mesh analysis and the use of supermesh.

J.live
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Homework Statement



About this circuit:

http://oi54.tinypic.com/eqwktd.jpg

If I were to find ix. Would I be using supermesh?


Homework Equations





The Attempt at a Solution

 
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No. You would just write three loop equations with three unknowns. Mesh analysis wants voltages, and that is a dependent voltage source that relies on a current. You traditionally need to use supermesh when you need to work around a current source (either dependent or independent).
 
In the last mesh, aren't i3 and ix separate currents ? Or ix= i3?
 
J.live said:
In the last mesh, aren't i3 and ix separate currents ? Or ix= i3?

They are different currents, but I_x can be written in terms of your mesh currents. Just sum the currents entering and exiting that node:

I_2 -I_3 - I_X = 0 \to I_X = I_2-I_3
 

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